CENTER OF MASS AND MOMENT OF INERTIA 159
Let F, Fig. 90, be the axis with respect to which v it is desired to find the moment of inertia of the cylinder. Let dly denote the moment of inertia of an element bounded by two transverse sections relative to the F-axis, and dly,, denote the moment of inertia of the same element relative to the
FIG. 90.
F"-axis, a parallel axis through the center of mass of the element. Then, by theorem II, we have dly = dlyť+ (x2+z2)dm,
where (x2 + z2} is the square of the distance between the two axes. Similarly the moment of inertia of the element about the F'-axis which is parallel to the F-axis and intersects the same elements of the cylinder, is given by
dly> = dlyť + z2 dm. Eliminating dlyť between the last two equations we obtain
dly = dly' + x2 dm = Ki2 dm + x2 dm,